Avalanche Photodetectors (Carrier multiplication simulation)


#1

Is it possible to simulate Avalanche PDs in Device. How can I simulate carrier multiplication? any suggestion on how this can be done?
How should I proceed? the carrier generation object in FDTD doesn’t consider the multiplication.
I am trying to replicate the results presented in the attached paper. A high-responsivity photodetector absent metalgermanium.pdf (2.4 MB)

Does Device calculate the fringe electric field between the two electrodes?

Thanks,
-A


#2

Simulation of APDs can be done with DEVICE CHARGE with some limitations. For example, you can use the solver to calculate the breakdown voltage of the APD. However, once breakdown is achieved, the current flow increases so rapidly that the solver fails to converge.

The carrier multiplication is modeled in the electrical solver (DEVICE CHARGE) and not in FDTD Solutions.

DEVICE has a built-in Poisson solver so it is able to incorporate the effects of fringe electric fields between electrodes.


#3

@aalam Thank you for the clarifications.
I think there might have been a mistake in the VPD example in the knowledge base, the responsivity of the VPD is shown to be above 1.3 A/W. This means the quantum efficiency is higher than 1, which doesn’t make sense.
https://kb.lumerical.com/en/pic_photodetectors_vertical_photodetector.html
The reference paper for the example says the responsivity is 0.9 A/W. Any thought on this?

Edit: in the VPD example the Device files already has an optical generation rate imported, however when I visualize the generation rate it is different from what is produced by FTDT. When the FDTD simulation is complete and the generation rate analysis is performed the results are different than what is in the DEVICE file, and the results don’t match.


#4

Hi @alireza. Thank you for bringing this to my attention. I will look into the mismatch in the responsivity value and the difference in generation rate. However, please note that the responsivity of the VPD is defined as ampere per watt. A value of 1.3 A/W means that the detector produces 1.3 A of photocurrent for an input optical power of 1 Watt. It does not imply that the quantum efficiency of the detector is greater than 1.


#5

Hi @aalam. Thank you. The generation rate script in FDTD assumes one electro-hole pair generation per 1 photon incident. That means the quantum efficiency is assumed to be 1 with is wrong for Silicon-Germanium. The reported quantum efficiency for Ge is between 0.6-0.8.
On the other hand even we assume perfect quantum efficiency of 1, the responsivity cannot be above 1.25 A/W theoretically for 1550 nm wavelength (its clear if you look at responsivity equation based on quantum efficiency). This means even an ideal PD can not produce 1.3 A of photocurrent with an optical power of 1 Watts.


#6

Hi @alireza. You are right. A quantum efficiency of 1 makes the responsivity of the PD approximately 1.25 A/W assuming 100% absorption. The answer might be in the generation rate as you have mentioned the loaded generation rate is different from the one you get from FDTD. I’ll look into it and let you know. Thanks.


#7

There are a few things going on here. First of all, you will notice that the length of the Ge slab in FDTD is 10 micron. However, in the DEVICE file, norm length is 50 micron which implies a 50 micron length of the detector in the y direction. So the current that you are getting from the DEVICE simulation is actually 5 times larger than what the structure from the FDTD file would give you. This means that if we assume an input power of 0.65 mW in the FDTD file, the equivalent input optical power for the DEVICE simulation is 5x0.65 mW = 3.25 mW. Using this value instead of the 0.65 mW in the script gives us a responsivity of approximately 0.27 A/W. Far below the theoretical maximum.

Now this value is less than the reported value in the paper as well. This is most likely due to a mismatch in the detector dimension and input optical power values. I will have to look into that in more details.


#8

@aalam Thanks Ahsan. Does the norm length mean the results are extrapolated in one direction for 50 microns? or both directions for 25 microns for a total of 50 um?
I think using a 2D-Y normal solver is not appropriate for this case, since the generation rate changes in the Y direction and is not uniform. However when I change to 3D model I get an error: (No area for degenerate linear planar polygon
I18N Runtime Warning:
Missing ICU data file detected while processing directory containing libut.{dll|so|sl|dylib}.)


#9

Hi aalam
I also have a question about the 3dB bandwidth calculation of the VPD example in you website
as we all know,3dB bandwidth change with different area of the PD,but when I use the scirpt of 3dB calculation in your example,I found the 3dB bandwidth always be the same while the area of PD is different.
can you help me how to solve this question.Thanks!


#10

Please see this discussion here for the question about effect of length on bandwidth: Selecting Parameter in Transient Simulation in DEVICE.

For the error message (which is arising because the solver is failing to create the 3D geometry) please see this post: Geometry building issue in 3D simulation for DEVICE.


#11

A post was split to a new topic: Error: no area for degenerate linear polygons


#12

I am using “Grad Phi” to simulate the impact ionization in InGaAs/InP planar APDs and the simulated breakdown voltage and the dark current matches closely my experimental data. As I have understood from various posts here, “Grad Phi” is not a valid “driving force” to simulate the heterostructure APDs.
My question is “why grad phi is not valid for heterostructure APDs?” Even though, InGaAs/InP is a heterostructure, but the charge sheet and the multiplication region are both located in the InP.


#13

The “grad phi” option calculates the electric field from the gradient of the quasi Fermi level which can give a wrong (high) value for electric fields at heterojunctions. This high electric field can then intorduce breakdown at the heterojunction even if the actual breakdown is supposed to happen somewhere else in the device. In your case I am guessing that the electric field it is calculating at the heterojunction is not large enough to effect the location of the breakdown and therefore you are getting the right behavior from the simulation.